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Math Help - Splitting Fields - Dummit and Foote - Exercise 1, page 545

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    Super Member Bernhard's Avatar
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    Splitting Fields - Dummit and Foote - Exercise 1, page 545

    Dummit and Foote Exercise 1 on page 545 reads as follows:

    Determine the splitting field and its degree over  \mathbb{Q} for  x^4 - 2 .

    ------------------------------------------------------------------------------------------------------

    I have started on the solution to this exercise as follows:

    The two roots of  x^4 - 2 are  \alpha = \sqrt[4]{2} and  \beta = \sqrt[4]{2}i

    Thus the splitting field is    \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i )

    We note that    \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i ) =    \mathbb{Q}(\sqrt[4]{2}, i ) since the product of   \sqrt[4]{2} and i must be in   \mathbb{Q}(\sqrt[4]{2}, i )


    The element  \alpha = \sqrt[4]{2} \in \mathbb{R} is algebraic over  \mathbb{Q} and so  m_{\alpha}(x) = m_{\sqrt[4]{2}}(x) = x^4 - 2 is the minimal polynomial for  \alpha = \sqrt[4]{2} over   \mathbb{Q}  .

    Thus the degree of  \alpha = \sqrt[4]{2} over   \mathbb{Q}  is the degree of  m_{\alpha}(x) = 4

    Hence  [\mathbb{Q}(\sqrt[4]{2}) \ : \ \mathbb{Q}] = 4



    The element  \beta = \sqrt[4]{2}i  \in \mathbb{C} is also algebraic over   \mathbb{Q}  and so  m_{\beta}(x) = m_{\sqrt[4]{2}i}(x) = x^4 - 2 is the minimal polynomial for  \beta = \sqrt[4]{2}i over   \mathbb{Q}   .

    Thus the degree of  \beta = \sqrt[4]{2}i over     \mathbb{Q}  is the degree of  m_{\beta}(x) = 4

    Hence  [\mathbb{Q}(\sqrt[4]{2}i) \ : \ \mathbb{Q}] = 4


    But where to from here - need to find the degree of the splitting field.

    Can someone please confirm that my reasoning above is valid and show me the way forward from here?

    Peter
    Last edited by Bernhard; October 2nd 2013 at 02:02 AM.
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